# Question 66a05

Mar 20, 2015

The partial pressure of Ar will be $\text{0.734 atm}$.

So, you know that you start with a certain mass of argon in the flask. The first thing you need to do is figure out exactly how many moles of argon you have

$\text{1.20 g Ar" * "1 mole Ar"/"40.0 g" = "0.0300 moles Ar}$

Now you add an unknown number of moles of ethane vapor to the flask. Since you know volume, total pressure, and temperature, you can use the ideal gas law equation to figure out how many moles of both argon and ethane can be found in the flask

PV = nRT => n_("total") = (P_("total")V)/(RT)

n_("total") = ("1.250 atm" * "1.00 L")/(0.082("atm" * "L")/("mol" * "K") * (273.15 + 25)"K") = "0.0511 moles"

So, your mixture contains a total of 0.0511 moles of gas, argon and ethane. Now, you can express the partial pressure of argon by using its mole fraction and the total pressure in the flask

${P}_{\text{argon") = chi_("argon") * P_("total}}$, where

${P}_{\text{argon}}$ - the partial pressure of argon;
${\chi}_{\text{argon}}$ - its mole fraction, i.e. the ratio between the number of moles of argon and the total number of moles present in the mixture;
${P}_{\text{total}}$ - the total pressure in the flask;

Since you know how many moles of argon you have, its partial pressure will be

P_("argon") = "0.0300 moles Ar"/"0.0511 moles gas" * "1.250 atm"

P_("argon") = "0.734 atm"#